=ua−ub ∆u=ub−u mgh when is positive (block falling),...

Electric Potential Energy

h

Think of gravitational potential energy.

When the block is moved vertically up

against gravity, the gravitational force does

negative work (you do positive work), and the

potential energy (U) increases.

When the block falls vertically down, the

gravitational force does positive work, and

the potential energy (U) decreases.

a

b

PHYS 153 08W 1PHYS 153 08W 1PHYS 153 08W 1

baab UUW −=mghUUU ab =−=∆In this case,

When is positive (block falling), baW → ba UU > , and is negative.U∆

The potential energy decreases.

When the block is lifted up, the gravitational work is negative, and

is positive.

U∆

Gravitational force is a conservative force.

The work done by a conservative force has 4 properties.

1. It is reversible.2. It is independent of the path between the start and end points.

3. When the start and end points are the same, total work is zero.

4. The work can be expressed as a difference in potential energy.

The electric field force is conservative, and the ideas above apply to the work

done in moving a charge in an electric field.

For a positive charge moving

in the direction of E, U

decreases, and moving

opposite to E, U increases.


Fig. 23.3

opposite to E, U increases.

The opposite is true for a

negative charge.

For a conservative force, the

change in K.E. is minus the

change in P.E.

)( abab UUKK −−=− bbaa UKUK +=+or

Can easily show that because the electrostatic force is conservative, the work

done by this force is independent of path.

When a force acts on a particle that moves from point a to point b, the

work done is given by a line integral,

→

F

∫∫ ==→→

→

b

a

b

aba dlFldFW φcos.

→

ld is the incremental displacement along the particle’s path, and φis the angle between

→

F and →

ld at any point.


is the angle between F and ld at any point.

Fig. 23.6

In the figure, a and b do not lie on the same

radial line.

∫ ∫==→

b

a

b

aba EdrqdlFW 0cosφ

since drdl =φcos

So, the work done during a small

displacement →

ld depends only on the change dr in the radial distance r,

independent of the path.

(a) Electric potential energy in a uniform field.

+

+

+

+

+

-

-

-

-

-

•q0 →

F

dWork done over some distance d is

EdqFdW 0==

Work done by the field is positive, so

the potential energy of the particle


+

+

-

-

x0

the potential energy of the particle

decreases.

In going from a to b, the decrease in

a b00ExqP.E. is

(b) Electrical P.E. of two point charges.

Consider a positive test charge q0 a distance r from a positive charge q.

Here the force is not constant with

distance as in the uniform field.

If the force moves the particle from

a to b along a radial line

drr

qqrdFW

b

a

b

aba ∫∫ ==

→→

→ 2

0

04

1.

πε

)11

(4

0

rr

qqUW −=∆−=

πε


)(4 0 ba rr

UW −=∆−=πε

The work, and hence change in P.E., depends

only on the end points. Here the work is positive

and the P.E. change is negative.

Recall, the work done does not depend on the path

between the end points.

Suppose we have infinity as our reference point, where U=0 and bring q0from infinity to point a.

Fig. 23.5

Then

ar

qqU 0

04

1

πε= )0( =∞U

Thus, the potential energy U of a test charge q0 at any distance r from charge

q is

r

qqU 0

04

1

πε= (electric potential energy of two point charges

q and q0.)

(c) Electric potential energy of several point charges.


(c) Electric potential energy of several point charges.

Fig. 23.8

Recall the resultant electric field is the

superposition of all individual fields. The

work done to move q0 to point a is the sum

of the work done against each field. So the

P.E. of charge q0 at point a is the algebraic

sum of the potentials of each pair of charges

involving q0.

∑=+++=i i

ia

r

qq

r

q

r

q

r

qqU

0

0

3

3

2

2

1

1

0

0

4...)(

4 πεπε

If q0 is moved to another point b, then Ub is given by the same expression

with the distances ri measured to point b. The work done in moving q0 from

a to b is ba UU −

The work done in bringing together all the charges in Fig. 23.8 (not just q0)

is given by

∑qq1


∑<

=ji ij

ji

r

qqU

04

1

πε(i<j so that we include each pair only once)

Electric Potential

Potential is the potential energy per unit charge. The symbol is V, and the

unit is the volt.

0q

UV = V, or J/C.

Refer back to Fig. 23.3 on slide 2, and use the equations on slide 1.

The work done by the field in moving the positive charge from a to b,

baab UUW −=

The electric field does positive work, and the potential energy decreases.

The potential is given by abbabaab VVVq

U

q

U

q

W=−=−=

000

The work done by the field in moving the charge from a to b is positive,

And so Va > Vb.


(a) Potential of a single point charge.

Recall the expression for the electric potential energy of two point

charges, q and q0.

Hence r

q

q

UV

00 4

1

πε==

What is the reference point here?

(b) Potential due to a collection of point charges.

Recall the expression for the electric potential energy for a point charge q0and a collection of charges qi.

Hence ∑==i i

i

r

q

q

UV

00 4

1

πεPotential at point a a distance ri from qi.

Potential here is calculated with respect to the same point in (a). This point is infinity, where

V=0.

(c) Potential due to a continuous distribution of charge.


(c) Potential due to a continuous distribution of charge.

∫=r

dqV

04

1

πεr is the distance from dq to the point where V is being

evaluated, again with respect to inifinity.

•q0Recall, potential is the work per

unit charge.

Suppose a positive charge q0 is

moved from a to b.

a

b→→

→ ∫= ldEqWb

aba .0

∫∫ ==−∴→→ b

a

b

aba dlEldEVV φcos.

In this case is negative, and hence Vb > Va .baW →

We see from this that the units of E can be V/m as well as N/C.

PHYS 153 08W 10PHYS 153 08W 10

Problem 22.48

A solid conducting sphere with raduis R carries positive total charge Q

The sphere is surrounded by an insulating shell with inner radius R and

outer radius 2R. The insulating shell has a uniform charge density .

(a) Find the value of so that the net charge of the entire system is zero.

(b) If has the value found in part (a), find the magnitude and direction of

electric field in each of the regions 0 < r < R, R < r < 2R, r > 2R. Plot a

graph of the radial component of E as a function of r.

(c) Do the results agree with the general rule that the electric field is

discontinuous only at locations where there is a thin sheet of charge?

ρρ

ρ

PHYS 153 08W 11PHYS 153 08W 11

Examples of calculating Electric Potential

Problem 23.32

A total electric charge of 3.50 nC is distributed uniformly over the surface of a

metal sphere with a radius of 24.0 cm. If the potential is zero at a point at

infinity, find the value of the potential at the following distances from the centre

of the sphere:

(a) 48.0 cm, (b) 24.0 cm, (c) 12.0 cm.

Recall that the electric field due to a uniformly

charged sphere is given by

PHYS 153 08W 12

Fig. 23.17

charged sphere is given by

2

04

1

r

qE

πε= (same as for a point charge at the

centre).

The potential due to a point charge is

r

qV

04

1

πε=

Problem 23.33

A uniformly charge thin ring has radius 15.0 cm and a total charge of 24.0 nC.

An electron is placed on the ring’s axis a distance of 30.0 cm from the centre

of the ring and is constrained to stay on the axis of the ring. The electron is then

released from rest.

(a) Describe the subsequent motion of the electron.

(b) Find the speed of the electron when it reaches the centre of the ring.

PHYS 153 08W 13

Problem 23.34

An infinitely long line of charge has linear charge density 5.00x10-12 C/m.

A proton (mass 1.67x10-27 kg. and charge 1.60x10-19 C) is 18.0 cm from

the line and moving directly toward the line at 1.50x103 m/s.

(a) Calculate the proton’s initial K.E.

(b) How close does the proton get to the line of charge?

PHYS 153 08W 14

=ua−ub ∆u=ub−u mgh when is positive (block falling),...

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